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So I am given following differential equation: $$ \dfrac{dy}{dt} = \frac{1}{(y-1)(y+2)} $$ and I am asked to plot how $y(t)$ vs $t$ graph looks like for $y(0)= 0$.

My attempt: First I can see that no values of $y$ will make $\frac{dy}{dt} = 0$, so there is no equilibrium solution. I have discontinuities at $1, -2$. But after drawing phase line I can see that for values of $y$ ranging from $-2$ and $1$, $\frac{dy}{dt} <0$. If I were to plot the graph of $y(t)$ vs $t$ for the Initial Value Problem, $y(0)=0$, how would it look like?

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Noting that $\frac{dy}{dt} < 0$ for $-2 < y < 1$, you're on the right track. Note that outside of this range, $\frac{dy}{dt}$. With a gradient that changes sign upon crossing $y=-2$ and $y=1$ suggests that the discontinuities of $\frac{dy}{dt}$ may be points where the gradient becomes infinite, i.e. the slope is vertical. This, along with the fact that a plot of $y$ against $t$ must go through the origin according to the boundary condition, gives enough information to produce a rough sketch, such as:

enter image description here

The assumption that $y(t)$ is continuous is made. To confirm this sketch, it would be best to solve the differential equation. Note the differential equation can be rewritten as:

$$\int{(y-1)(y+2)\,dy}=\int{t\, dt}$$

$$\therefore \frac{1}{3}y^3 + \frac{1}{2}y^2 -2y = t + const.$$

Applying the boundary condition, it can be seen that $t=t(y)$ is a cubic:

$$t=\frac{1}{3}y^3 + \frac{1}{2}y^2 -2y$$

And so $y=y(t)$ will be a cubic mirrored about the line (y=t), and its plot will look as such (The horizontal axis should be labelled "$t$" and not "$x$"):

enter image description here

This corresponds to the sketch made earlier.

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    Thanks for the answer. Just a quick question. Shouldn't there be a bound on t for the given initial value problem so that the y stays between 1 and -2? – 2017-02-05
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    Why do you think there would be a bound on $t$? You haven't specified a bound on $y$ or $t$ in your question. – 2017-02-11